Solutions to Integration by Partial Fractions

SOLUTIONS TO INTEGRATION BY PARTIAL FRACTIONS

SOLUTION 9 : Integrate $\displaystyle{ \int {x^2 - x + 1 \over (x+1)^3 } \,dx }$ . Decompose into partial fractions (There is a repeated linear factor !), getting

$\displaystyle{ \int {x^2 - x + 1 \over (x+1)^3 } \,dx} = \displaystyle{ \int{\Big( {A \over x+1} + {B \over (x+1)^2} + {C \over (x+1)^3}\Big)}\,dx}$

(After getting a common denominator, adding fractions, and equating numerators, it follows that $\ \ A(x+1)^2 + B(x+1) + C = x^2 - x + 1$ ;
let $\displaystyle{x = -1: \ A(0) + B(0) + C = 3 \longrightarrow C=3}$ ;
let $\displaystyle{x = 0: \ A(1) + B(1) + C = A + B + 3 = 1 \longrightarrow A + B = -2 }$ ;
let $\displaystyle{x = 1: \ A(4) + B(2) + C = 4A + 2B + 3 = 1 \longrightarrow 2A + B = -1}$ ;
it follows that $\ B = -3$ and .)

$= \displaystyle{ \int{\Big( {1 \over x+1 } + { -3 \over (x+1)^2 } + {3 \over (x+1)^3 }\Big)} \,dx}$

$= \displaystyle{ \int \Big( {1 \over x+1 } - 3 (x+1)^{-2} + 3 (x+1)^{-3} \Big) \,dx}$

$= \displaystyle{ \ln \vert x+1\vert - 3 { (x+1)^{-1} \over (-1)} + 3 { (x+1)^{-2} \over (-2)} } + C$

$= \displaystyle{ \ln \vert x+1\vert + { 3 \over x+1 } - { 3 \over 2(x+1)^2 } + C }$ .

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SOLUTION 10 : Integrate $\displaystyle{ \int {x^3 + 4 \over (x^2 -1) (x^2 + 3x + 2) } \,dx }$ . Factor and decompose into partial fractions (There is a repeated linear factor !), getting

$\displaystyle{ \int {x^3 + 4 \over (x^2 -1) (x^2 + 3x + 2) } \,dx } = \displaystyle{ \int {x^3 + 4 \over (x+1)(x -1) (x + 2)(x+1) } \,dx }$

$= \displaystyle{ \int {x^3 + 4 \over (x+1)^2 (x -1) (x + 2) } \,dx }$

$= \displaystyle{ \int {\Big({A \over x+1} + {B \over (x+1)^2} + {C \over x-1} + {D \over x+2 } \Big)} \,dx }$

(After getting a common denominator, adding fractions, and equating numerators, it follows that

$\ \ A(x+1)(x-1)(x+2) + B(x-1)(x+2)$ ;

let $\displaystyle{x = -1: \ A(0) + B(-2) + C(0) + D(0) = 3 \longrightarrow B = -{3\over 2}}$ ;
let $\displaystyle{x = 1: \ A(0) + B(0) + C(12) + D(0) = 5 \longrightarrow C ={5\over 12}}$ ;
let $\displaystyle{x = -2: \ A(0) + B(0) + C(0) + D(-3) = -4 \longrightarrow D ={4\over 3}}$ ;
let $x = 0: \ A(-2) + B(-2) + C(2) + D(-1) = -2A + 3 + \displaystyle{5 \over 6} - \displaystyle{4 \over 3} = 4$
$\longrightarrow A = -\displaystyle{3\over 4}$ .)

$= \displaystyle{ \int{ \Big( {-3/4 \over x+1} + {-3/2 \over (x+1)^2} + {5/12 \over x-1} + {4/3 \over x+2} \Big)}\,dx}$

$= \displaystyle{ \int{ \Big( -(3/4){1 \over x+1} - (3/2) (x+1)^{-2} + (5/12){1 \over x-1} + (4/3) {1 \over x+2} \Big) } \,dx}$

$= \displaystyle{ -(3/4) \ln \vert x+1\vert - (3/2) {(x+1)^{-1} \over (-1) } + (5/12) \ln \vert x-1\vert + (4/3) \ln{\vert x+2\vert} + C}$

$= \displaystyle{ -{3 \over 4} \ln \vert x+1\vert + { 3\over 2(x+1) } + {5 \over 12} \ln{\vert x-1\vert} + {4 \over 3} \ln{\vert x+2\vert}+ C }$ .

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SOLUTION 11 : Integrate $\displaystyle{ \int { x^3 + 2x - 1 \over (x^2 - x -2)^2 } \,dx }$ . Factor and decompose into partial fractions (There are two repeated linear factors !), getting

$\displaystyle{ \int { x^3 + 2x - 1 \over (x^2 - x -2)^2 } \,dx } = \displaystyle{ \int { x^3 + 2x - 1 \over ((x+1) (x-2))^2 } \,dx }$

$= \displaystyle{ \int { x^3 + 2x - 1 \over (x+1)^2 (x-2)^2 } \,dx }$

$= \displaystyle{ \int { \Big( {A \over x+1 } + { B \over (x+1)^2 } + {C \over x-2 } + { D \over (x-2)^2 } \Big) } \,dx }$

(After getting a common denominator, adding fractions, and equating numerators, it follows that

$\ \ A(x+1)(x-2)^2 + B(x-2)^2 + C(x+1)^2(x-2) + D(x+1)^2$ ;

let $\displaystyle{x = -1: \ A(0) + B(9) + C(0) + D(0) = -4 \longrightarrow B =-{4\over 9}}$ ;
let $\displaystyle{x = 2: \ A(0) + B(0) + C(0) + D(9) = 11 \longrightarrow D ={11\over 9} }$ ;
let $\displaystyle{x = 0: \ A(4) + B(4) + C(-2) + D(1) = 4A - {16 \over 9} - 2C + {11 \over 9} = -1 }$
$\displaystyle{ \ \ \ \ \ \ \ \ \longrightarrow 2A-C = -{2 \over 9} }$ ;
let $\displaystyle{x = 1: \ A(2) + B(1) + C(-4) + D(4) = 2A - {4 \over 9} - 4C + {44 \over 9} = 2 }$
$\displaystyle{ \ \ \ \ \ \ \ \ \longrightarrow A-2C = -{11 \over 9} }$ ;
it follows that $\ A = \displaystyle{7 \over 27}$ and $C = \displaystyle{20 \over 27}$ .)

$= \displaystyle{ \int{ \Big( { 7/27 \over x+1} +{-4/9 \over (x+1)^2} + { 20/27 \over x-2} + {11/9 \over (x-2)^2} \Big) } \,dx}$

$= \displaystyle{ \int{ \Big( (7/27){ 1 \over x+1} - (4/9) (x+1)^{-2} + (20/27){ 1 \over x-2} + (11/9) (x-2)^{-2} \Big) }\,dx}$

$= \displaystyle{ (7/27) \ln \vert x+1\vert - (4/9) { (x+1)^{-1} \over (-1) } + (20/27) \ln \vert x-2\vert + (11/9) { (x-2)^{-1} \over (-1) } + C }$

$= \displaystyle{ {7 \over 27} \ln \vert x+1\vert + { 4 \over 9(x+1) } + {20 \over 27} \ln \vert x-2\vert - { 11 \over 9(x-2) } } + C$ .

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SOLUTION 12 : Integrate $\displaystyle{ \int{ \sec^2 x \over \tan^3 x - \tan^2 x } \,dx }$ . Use the method of u-substitution first. Let

$u = \tan x$

so that

$du = \sec^2 x \ dx$ .

Substitute into the original problem, replacing all forms of , getting

$\displaystyle{ \int { \sec^2 x \over \tan^3 x - \tan^2 x } \,dx } = \displaystyle{ \int { 1 \over \tan^3 x - \tan^2 x } \, \sec^2 x \ dx }$

$= \displaystyle{ \int { 1 \over u^3 - u^2} \, du }$

(Factor and decompose the function into partial fractions. There is a repeated linear factor !)

$= \displaystyle{\int { 1 \over u^2 (u - 1) } \, du}$

$= \displaystyle{ \int { \Big({A \over u} + {B \over u^2 } + { C \over u - 1} \Big) } \, du}$

(After getting a common denominator, adding fractions, and equating numerators, it follows that $\ \ Au(u-1) + B(u-1) + C u^2 = 1$ ;
let $\displaystyle{u = 0: \ A(0) + B(-1) + C(0) = 1 \longrightarrow B = -1 }$ ;
let $\displaystyle{u = 1: \ A(0) + B(0) + C(1) = 1 \longrightarrow C = 1 }$ ;
let $\displaystyle{u = -1: \ A(2) + B(-2) + C(1) = 2A + 2 + 1 = 1 \longrightarrow A = -1 }$ .)

$= \displaystyle{ \int{ \Big({-1 \over u} + {-1 \over u^2 } + {1 \over u-1} \Big) }\,du}$

$= \displaystyle{ \int{ \Big(-{1\over u} - u^{-2} + {1 \over u-1} \Big) }\,du}$

$= \displaystyle{ -\ln \vert u\vert - { u^{-1} \over (-1) } + \ln \vert u-1\vert + C }$

$= \displaystyle{ -\ln \vert\tan x\vert + { 1 \over \tan x } + \ln \vert\tan x -1\vert + C }$

$= \displaystyle{ \cot x + \ln \vert\tan x -1\vert - \ln \vert\tan x\vert + C }$

(Recall that $\ \ln m - \ln n = \ln \Big( \displaystyle{ m \over n } \Big)$ .)

$= \displaystyle{ \cot x + \ln { \vert\tan x - 1\vert \over \vert\tan x\vert } + C }$

$= \displaystyle{ \cot x + \ln \Big\vert { { \tan x \over \tan x } -{ 1 \over \tan x } } \Big\vert + C }$

$= \displaystyle{ \cot x + \ln \vert 1 - \cot x \vert + C }$ .

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SOLUTION 13 : Integrate $\displaystyle{ \int {x^3 + 8 \over (x^2 -1) (x-2) } \,dx }$ . Because the degree of the numerator is not less than the degree of the denominator, we must first do polynomial division. Then factor and decompose into partial fractions, getting

$\displaystyle{ \int {x^3 + 8 \over (x^2 -1) (x-2) } \,dx } = \displaystyle{ \int {x^3 + 8 \over x^3 - 2x^2 - x + 2 } \,dx }$

$= \displaystyle{ \int { \Big( 1 + { 2x^2 + x +6 \over x^3 - 2x^2 - x + 2 } \Big) } \,dx }$

$= \displaystyle{ \int { \Big( 1 + { 2x^2 + x +6 \over (x+1)(x -1) (x-2)} \Big) } \,dx }$

$= \displaystyle{ \int { \Big( 1 + {A \over x+1} + {B \over x -1} + { C \over x-2} \Big) } \,dx }$

(After getting a common denominator, adding fractions, and equating numerators, it follows that

$\ \ A(x-1)(x-2) + B(x+1)(x-2) + C(x+1)(x-1)$ ;

let $\displaystyle{x = -1: \ A(6) + B(0) + C(0) = 7 \longrightarrow A ={7\over 6}}$ ;
let $\displaystyle{x = 1 : \ A(0) + B(-2) + C(0) = 9 \longrightarrow B =-{9\over 2}}$ ;
let $\displaystyle{x = 2: \ A(0) + B(0) + C(3) = 16 \longrightarrow C ={16\over 3}}$ .)

$= \displaystyle{ \int{ \Big( 1 + {7/6 \over x+1} + {-9/2 \over x -1} + { 16/3 \over x-2} \Big) } \, dx}$

$= \displaystyle{ \int{ \Big( 1 + (7/6){1 \over x+1} - (9/2){1 \over x -1} + (16/3){1 \over x-2} \Big) } \, dx}$

$= \displaystyle{ x + {7 \over 6} \ln \vert x+1\vert - {9 \over 2} \ln \vert x-1\vert + {16\over 3} \ln \vert x-2\vert + C }$ .

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SOLUTION 14 : Integrate $\displaystyle{ \int { e^x \over (e^x -1) (e^x+3) } \,dx }$ . Use the method of u-substitution first. Let

so that

Substitute into the original problem, replacing all forms of , getting

$\displaystyle{ \int {e^x \over (e^x -1) (e^x+3)} \,dx } = \displaystyle{ \int {1 \over (e^x -1) (e^x+3)} \,e^x dx }$

$= \displaystyle{ \int { 1 \over (u -1) (u + 3)} \, du }$

(Decompose into partial fractions.)

$= \displaystyle{ \int{ \Big( {A \over u-1} + { B \over u+3 } \Big) } \,du }$

(After getting a common denominator, adding fractions, and equating numerators, it follows that $\ \ A(u+3)+ B(u-1) = 1$ ;
let $\displaystyle{u = 1 : \ A(4) + B(0) = 1 \longrightarrow A = {1\over 4}}$ ;
let $\displaystyle{u = -3 : \ A(0) + B(-4) = 1 \longrightarrow B = -{1\over 4}}$ .)

$= \displaystyle{ \int{ \Big( {1/4 \over u-1} + {-1/4 \over u+3} \Big)}\,dx}$

$= \displaystyle{ \int{ \Big( (1/4){1 \over u-1} - (1/4){1 \over u+3} \Big)} \,dx}$

$= \displaystyle{ {1\over 4} \ln \vert u-1\vert - {1 \over 4} \ln \vert u+3\vert + C }$

$= \displaystyle{ {1 \over 4} \ln \vert e^x-1\vert - {1 \over 4} \ln \vert e^x+3\vert + C }$

$= \displaystyle{ {1 \over 4} \Big( \ln \vert e^x-1\vert - \ln \vert e^x+3\vert \Big) + C }$

(Recall that $\ \ln m - \ln n = \ln \Big( \displaystyle{ m \over n } \Big)$ .)

$= \displaystyle{ {1\over 4} \ln { \vert e^x-1\vert \over \vert e^x+3\vert } + C }$ .

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About this document ...

Duane Kouba 2000-05-02